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Most likely a material like diamond dust or other non capacitive/inductive matrix.
As long as it equals or exceeds solder's thermal conductivity. There had to be a good technical reason for Intel to abandon soldered IHS, not just a few pennies per/cost savings. Maybe soldering the IB drastically reduces final yields due to the thermal stress of the soldering process on the tri-gate 22nm denser structure.

Its the only barrier left... delid to confirm what is causing/influencing the thermal buildup.. If it makes little difference, then it is because of the chip's density. But if it makes a very big difference, then Intel has some adjustments to make..

FWIW in the context it was given to me, the literal translation of "secret sauce" would be "we aren't going to tell you". It wasn't meant to actually imply anything about the nature of the interface material. I might have caused a little confusion with my humor in the previous post.

Personally, I have little grounds to believe anything, but I believe they haven't used solder on Ivy Bridge where we are seeing high temps... With a delid to eliminate the extra heat interface and the IHS barrier, and some modification to the socket area to allow good direct contact with the die, I think temps would be a lot better, like in the 15-20C better range.

However there is 47.3% more surface area with 22nm compared to SB so there is no cramming involved.

Mater of fact intel had so much surface area they shrunk the die from 216 mm≤ to 160 mm≤ and they are able to keep the same density because there is 47.3% more surface are switching to 22nm

Smaller die = less surface area. The die size is a measure of surface area.
They aren't talking about the amount of space a transistor takes up, they're talking about how many transistors are in a given die size.

"Two things are infinite: the universe and human stupidity; and I'm not sure about the the universe." -- Einstein (maybe)

Well, the problem is you are comparing two different systems... mobo's, ram, hdd, etc. So though it may not make up the difference, its tough to compare non like systems and come to a conclusion. That and, he mentioned stock you go to overclocked to prove your point... LOL!

Yes i know it's not the same system to compare, I was just showing diaz load overclocking watts on IB and my SB also ldle watts on IB so he will know how hot it can get.

Originally Posted by diaz

incorrect.. It is only hotter at the core where the temps are taken. The heatsink only has to deal with 77watts vs the 95 of SB for example. The heatsink should still be quite cool to the touch on most occasions with IB.

However since you brought up overclocking watts SB vs IB i wll be doing that when i purchase a IB.

Smaller die = less surface area. The die size is a measure of surface area.
They aren't talking about the amount of space a transistor takes up, they're talking about how many transistors are in a given die size.

Yes i agree and when you shrink to 22 nm you can fit more in the same space without cramming transistors, so the power density stays the same.

intel does not pack more transistors in the same space, there shrunk so you can fit more and they reduced the die size also some times because they have more surface area than needed, intel knows what there doing when it comes to transistor and tracers shrinking.

TDP is 77W for IB bobnova. 95W is marked on the boxes so that board manufacturers design to the 95W spec for Z77, otherwise they could design for 77W TDP, and then you'd have Z77 boards that weren't fully compatible with Sandy Bridge.